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Let a vertical tower AB have its end A on the level ground. Let C be the mid-pont of AB and P be a point on the ground such that AP=2AB. If $\angle$ BPC= β , then tanβ is equal to

Answer:

Explanation:

LET AB= h, then AD= 2h, and AC=BC= $\frac{h}{2}$

  Again , let $\angle$ CPA= α

 Now. in ΔABP

   $\tan(\alpha+\beta)=\frac{AB}{AP}=\frac{h}{2h}=\frac{1}{2}$

Also, in Δ ACP, $\tan\alpha=\frac{AC}{AP}=\frac{\frac{h}{2}}{2h}=\frac{1}{4}$

Now, $\tan\beta=\tan[(\alpha+\beta)-\alpha]$

                   $=\frac{\tan(\alpha+\beta)-\tan\alpha}{1+\tan(\alpha+\beta)\tan\alpha}$

$=\frac{\frac{1}{2}-\frac{1}{4}}{1+\frac{1}{2}\times\frac{1}{4}}=\frac{\frac{1}{4}}{\frac{9}{8}}=\frac{2}{9}$